SET FLASK_APP在Anaconda中不起作用，只有在使用 app.py 时才能运行

Question

我正在激活一个conda环境，设置FLASK_APP = flaskblog.py - 可以看到运行"SET"命令时设置的环境变量，但在Windows 10上收到此错误。

(ariel) C:flask_blog>SET FLASK_APP = flaskblog.py
(ariel) C:flask_blog>flask run
* Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
Use a production WSGI server instead.
* Debug mode: off
Usage: flask run [OPTIONS]
Error: Could not locate a Flask application. You did not provide the "FLASK_APP" environment variable, and a "wsgi.py" or "app.py" module was not found in the current directory.
(ariel) C:flask_blog>dir
Volume in drive C is OSDisk
Volume Serial Number is 36E9-84F4
Directory of C:flask_blog
08/01/2020  12:57    <DIR>          .
08/01/2020  12:57    <DIR>          ..
08/01/2020  12:45    <DIR>          .vscode
08/01/2020  12:50               105 flaskblog.py
08/01/2020  12:55    <DIR>          __pycache__
1 File(s)            105 bytes
4 Dir(s)  298,386,743,296 bytes free

我可以让应用程序运行的唯一方法是将 flaskblog.py 重命名为 app.py。

我从Anaconda提示符运行，而不是PowerShell。

建议赞赏。

Answer 1

啊....设置环境变量时不要使用空格。

SET FLASK_APP = flaskblog.py

应该是

SET FLASK_APP=flaskblog.py

Answer 2

我认为这是因为您没有为flask_app设置 ENV

如果您使用PowerShell，请使用此

$env:FLASK_APP = "hello"
flask run

如果您使用cmd，请使用此

set FLASK_APP=hello
flask run

有关更多细节，请查看文档

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